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Would it be a good idea to use dihedral on a biplane?
My engineering class uses model plane kits to build "frankenflyers" which use parts from multiple model planes to make one plane that will carry a maximum load (pennies). I was wondering what the best configuration would be. We have three wings and three bodies. I was thinking we should use a biplane for added lift, but I couldn't find much on dihedral wings being used on a biplane. Open to criticism and ideas.

If your aircraft is meant to be a "free flight" model, meaning that it flies without any guidance from any pilot, then generous dihedral will help keep the wings level.
Even if your aircraft is meant to fly under radio control, so long as the mission is load-carrying rather than aerobatics, some dihedral will help ease the pilot workload.  You may want to mix some rudder to your aileron channel to help it turn better, otherwise the dihedral will tend to create some resistance to aileron roll/turn inputs.  If you don't have a rudder, don't use very much dihedral.  If you have only a rudder and no ailerons, use lots of dihedral.
Many full-scale (piloted) biplanes do in fact have dihedral on one or both sets of wings.  Most typically the lower wing has dihedral while the upper wing is straight, because that helps improve ground clearance of the lower wingtips, but in some cases both sets of wings have equal or nearly equal dihedral.  (See for example this 3-view diagram of the "Bristol Fighter".  The combination of a straight lower wing and a dihedral upper wing has never been used before, to the best of my knowledge-- here's your chance to be the first!

Actually I think sometime in the 20's or 30's there was a biplane design, possibly British, possibly a torpedo bomber, that had dihedral only on the upper wings.  I'll modify the answer and include a link if I (or someone else) come/s across a specific reference.

The Sopwith 1 1/2 Strutter had a straight lower wing and a mildly dihedral upper.  It's probably been done again in the intervening hundred-plus years...

Yes, probably so.

@ZeissIkon -- Yet in this particular case, when I hold a ruler up to this three-view, it doesn't seem to support your assertion-- https://en.wikipedia.org/wiki/Sopwith_1%C2%BD_Strutter#/media/File:Sopwith_1%C2%BD_Strutter_drawing.jpg

Yep, you're correct.  I was going by photos I found in a quick search, in which the lower wing does look flat.

@ZeissIkon-- got it--

PS actually I think the design stuck in my memory had a much LONGER lower wing than upper wing. Really bizarre.  I don't have a clear recollection of any design with dihedral on the upper wing only, but one probably does exist.

If you mean that you have three fuselages and three wings to choose from, and you want to maximize the weight of the payload carried, and you may use more than one wing, then build a triplane.  More wing means more lift!
As quietflyer answered, dihedral increases stability (as does sweepback, somewhat), and reduces aileron authority. But this is true for monoplanes and biplanes and triplanes and stranger things.  That's why you couldn't find much written about this, because mono-vs-biplane and amount of dihedral don't interact much.  Just space the wings about one chord apart for their full span.


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How much does the fuselage of an airplane contribute to yaw stabilization?
Obviously considering that (almost) every plane has a vertical stabilizer, probably not, but for a relatively long aircraft does the fuselage measurably contribute to the yaw stabilization relative to the vertical stabilizer?

Sometimes it's hard to tell where the fuselage stops and the fin begins-- https://en.m.wikipedia.org/wiki/File:Couzinet_Arc-en-Ciel.jpg

This is not a dumb question at all, and the short answer is not at all.
Since the fuselage must be balanced around the center of gravity (cg), part of it will be forward of the cg. This part contributes a destabilizing effect and since the fuselage is a slender object, most of its lift will be produced near the nose. This means that the whole fuselage, even if it is very long, is still destabilizing as a whole.
The center of pressure of a symmetric airfoil in subsonic speed is at its quarter point. This is classical two-dimensional aerodynamics. Once this airfoil is turned into a wing, its aspect ratio (ratio between span and chord) will influence the location of the center of pressure such that it moves forward (towards the leading edge) the smaller the aspect ratio becomes. In the extreme case of very long chord and almost no span, the center of pressure moves into the leading edge. For a fuselage this means that its neutral point is way ahead of its center of gravity.
The vertical stabilizer is closer to a wing and also more effective. The aspect ratio also influences the amount of lift that can be achieved with a given angle of attack (lift curve slope) and for slender bodies the lift curve slope per radian is as low as π/2 times the aspect ratio while for a wing of infinite aspect ratio it is 2π. This is all valid for attached flow, so with larger amounts of sideslip and flow separation on the leeward side things will change and the exposed area will become more important.
The heavy engines and very lightweight structures of early airplanes would allow a very short forward fuselage while the rear fuselage would have a large side area. Still, even those needed a small vertical tail, mostly for control, but also for stabilization. Note that all airships were laterally unstable and required constant attention by the person at the rudder controls to stay on course.

The Nieuport 11 (picture source) should illustrate the point about early airplanes nicely. If you take away the rudder, the fuselage will start to yaw off course until flow separation starts and limits the amount of yaw. However, at this point drag will already be punishingly high, so the rudder helps to keep the fuselage on course.

A chum from years ago used to fly a Fokker DR-1 that was owned by a little museum. It was a full size replica, powered with a Lycoming O-320 disguised under the round cowling, so without the gyro effects of a rotary, but in any case he said it was pretty much neutral to slightly negative in yaw and would sit wherever you pointed it, and could perform incredible high rate wings-level skidding turns that felt like you were in a centrifuge.  He said it was exhausting to fly, but it was easy to see why Von Richtofen liked it, with its seeming ability to point and shoot in almost any direction.

@JohnK This flat turning was a trick introduced by Oswald Boelcke in the Fokker E.III. The initial instability together with the fully deflected rudder built up the yaw rate so the plane would continue to spin around even when flying sideways. It was not very effective, though, because the high yaw rate in combination with the low rate of fire of their guns would only be good for scaring their pursuers.

I sometimes feel I need to make a fake account just so that I can give these kind of answers more upvotes.

@KennSebesta Please spare the effort. I have enough reputation already.

Yes, the fuselage contributes to yaw stability, as do the wings and in some cases the engine(s). However, the fuselage contributes in a negative way, it is unstabilising.
What percentage the fuselage contributes, depends of course on the shape of the aeroplane. The figure below is from my university book (paper version only) on flight dynamics, by prof. Gerlach, which shows the dimensionless yaw coefficient $C_n$ as a function of side slip angle $\beta$, for a wind tunnel model of the Fokker F27:

The four graphs indicate, from left top to bottom, the $C_n/\beta$ of:

The wing/fuselage combination.
The fuselage only.
The wing only.
The complete aeroplane model, the only graph showing the desired positive $C_n/\beta derivative$

Notice that the wing/fuselage combination is not a simple addition of wing and fuselage contribution, due to the wing/fuselage interference. The wing only contribution is ever so slightly positive.
For the two graphs with the fuselage, $C_n$ is negative for positive $\beta$, indicating a de-stabilising yaw coefficient as a function of slip angle. For the F27 the contribution of the fuselage/wing is about -30%, the vertical tail is an absolute necessity!
We can see that the straight wings of the F27 contribute very little to slip stability. Contribution to slip stability of the wings depends on the wing dihedral and the wing sweep angles. How the wing sweep angle contributes is intuitively clear from the figure below:

Ahem - isn't c$y$ the side force coefficient? For yaw stability the coefficient to use is c${n\beta}$. And I'm quite sure that the F-27 fuselage is destabilizing in yaw, even in Prof. Gerlach's book.

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How can a small ducted fan produce a large amount of thrust?
This 15.2 cm ducted fan by Schuebeler can produce 175 Newtons of static thrust. How is this possible in such a small diameter?
If this seems normal, then why and how can this ducted fan produce that much force?

When you pack too much thrust into a ducted fan it becomes a turbofan.

A 16 hp motor to make only 40lbs of thrust is not all that impressive.  A 16 hp motor driving a large prop would make about 50-60 lbs.  It's only real benefit in the motoglider case is compactness.

Thrust depends greatly on diameter, so your curiosity is reasonable.
Why:  A first-principles answer says that (static) thrust depends directly on the rate of mass flow through the duct.  Because the duct's area is small, the speed of air through the duct must be large, and thus what must also be large is the fan's (static) pitch speed and its RPM.
See https://www.grc.nasa.gov/www/k-12/airplane/thrsteq.html for the case $p_e = p_0$.
For 175 N thrust, Schuebeler says the fan turns at 20,000 rpm and needs 12 kW.  Those numbers are not vastly different from an unshrouded 15 cm propeller, or a small turbine (a JetCat P220 does 220 N at 725 mL/min, converted via kerosene's MJ/L to roughly 450 kW; there's much more waste heat in a turbine's exhaust than in a ducted fan's gradually warming battery pack).
How:  Tight manufacturing tolerances, so the blades don't vibrate from imbalance, so the blades don't deform or break at high rpm, and so not too much air leaks around the gap between the blade tips and the walls of the duct.  (The same applies to turbines.)

Oh thank you for the clarification. I also have another question from this. How much airflow must exist in order to provide this much force?

Static thrust = mass flow rate times exit velocity = (density of air) * (area of duct) * (speed squared), from https://www.grc.nasa.gov/www/k-12/VirtualAero/BottleRocket/airplane/thrsteq.html.  There you go!

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Why does the Antonov AN-225 have anhedral wings, but a dihedral tail?
I know why it has negative dihedral wings, but I do not really know why it is combined with a positive dihedral horizontal stabilizer. What are the benefits this configuration provides?

The reason the AN-225 has anhedral wings is explained here: https://aviation.stackexchange.com/questions/6285/why-does-an-225-have-anhedral-wings-though-it-is-a-cargo-aircraft
And the reason the horizontal stabilizer angle is different to the wings' angle is probably covered by this question: https://aviation.stackexchange.com/questions/36810/why-do-f16s-have-anhedral-tail-wings  (i.e. to keep the tail surfaces away from the wing wake)

@DaveGremlin: Thanks for pointing out that this is almost a dupe. The tail dihedral is most likely also to move the two vertical surfaces up while keeping the attachment point close to the center in order to reduce tail bending moments. Note that the AN-124 tail has no dihedral and sits at the same height.

The reason for the wing anhedral has been answered here long ago.
Tail anhedral or dihedral can have several reasons:

To avoid ground contact during rotation (when the aircraft pitches up in order to lift off the ground). This is needed with a swept tail, so tail dihedral can be found on many airliners. Clearly, the An-225 tail has ample distance to the ground, because the vertical tail surfaces extend for ⅓ of their span below the horizontal tail, so any ground contact can be ruled out even for an un-dihedraled tail.
To move the horizontal tail away from the wing wake. When you look at the tail of the An-124, you will see no dihedral and a mounting point quite similar to that of the An-225. But if the An-124 has no problem with an un-dihedraled tail, so should the An-225. This can also not be the reason.
To add some more vertical tail surface. This trick has been employed on the Handley-Page Victor or the F-4 Phantom II. On the An-225, the outer verticals would shield much of the horizontal tail and little additional side force will be created by the dihedraled horizontal tail. So this cannot be the reason, either.
For a H-tail: To move the attachment point of the vertical surfaces closer to their middle. Imagine the An-225 horizontal tail had no dihedral: Then the verticals would need to be mounted on the tips of the horizontal near their bottom (they need to be so high up in order to allow the An-225 to rotate, after all). This would mean that all side loads would act above the attachment point, producing a sizeable bending moment which has to be carried along the spar of the horizontal tail, making the outer tail structure a lot heavier. With a little dihedral, the attachment point can be moved further up, close to the middle, so the bending moments in the vertical and horizontal tails can be a lot lower.

Antonov has designed H-tails before (on the An-14, the An-22 Antei, the An-28 and the An-38), and in all cases the vertical had a sizeable part of its surface below the horizontal tail. The only design on this list which also had some tail dihedral is the An-14 from 1958, and here the attachment point is almost in the center of the vertical. So it looks like the Antonov design bureau applied a trick on the An-225 which they had known for at least 30 years before the An-225 took off for the first time.

Anhedral is used on high wing swept wing aircraft because the dihedral effect of the sweep PLUS the aerodynamic dihedral effect of the T wing-to-fuselage interface adds up to too much total dihedral effect.  The roll/yaw couple (rolling moment created by yaw) becomes much too strong.  The anhedral brings the total dihedral effect down to a desirable level.
The dihedral of the horizontal tail can be done for two reasons; ground clearance for the vertical tail surfaces, and the effective side area presented by the canted surfaces can be considered to be adding to the vertical fin area (a V tail takes this to the ultimate extreme).
In the case of the Antonov I will bet that the tail is like that mostly to provide a little extra vertical clearance for ground vehicles.

… and for rotation.

I'll bet that the original design was straight, and can imagine the design merry-go-round that must've gone on - Problem; Horiz tail straight, but vert fins too close to ground - solution; move them up so the semi-T is now an L.  Ooops, bending loads at base of fins increased, now we have to reinforce that intersection, more weight.  Then some bright light in the room said "why don't we just crank the stab spar a few degrees.  Having the rudders canted is no big deal.  Problem solved."   Someone probably got a Hero Of The Soviet Union award for that.

I think that line of thought had already been completed at Antonov when the An-14 was designed.


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What are the issues involved in converting a quadjet into a twinjet?
As you are probably all aware already, the fewer engines you have, the better your fuel efficiency. That’s why the airlines are selling off their quadjet 747s and buying twinjet 777s. It's why they’ve already sold off their trijet DC-10s, L-1011s, and MD-11s and replaced them with twinjet 767s, 777s, and 787s. It's why the quadjet A340 lost out to the twinjet A330, and the quadjet 747-8 and A380 are struggling to sell while the twinjet 777 and twinjet A350 are selling like 737s. It's why the trijet MD-11 went over like a lead balloon compared to the twinjet 767, 777, and A330.
One obvious solution to the kerosene-guzzling-quadjet problem would be consolidation: replacing the two engines on each wing1 with a single, bigger engine (for instance, replacing a 747’s four JT9Ds, CF6s, RB211s, or GEnxs with a pair of GE90s, or an A340’s four CFM56s or Trent 500s with a pair of Trent 700s or GE90s).2  However, this generates the obvious problem that the wing might need to be redesigned for the different stress distribution of a twinjet, as opposed to a quadjet, design.  Still, the change doesn’t have to be large.  For instance, the A340’s wing is already almost identical to the A330’s; just deactivate the extra fuel tanks, chop off the bulge under the wing that was introduced to keep the airflow around the outboard engine from breaking the wing off, take away the outboard pylons, swap out the inboards for a pair of A330 engines, and you’re in business.
Yet, despite the massive efficiency advantage to be gained, there’ve been hardly any serious proposals to reengine quadjets into twinjets in this sort of way, indicating that there’re other issues that’re showstoppers for twinjetting a quadjet; what are they?

1: I’m leaving trijets (DC-10, L-1011, MD-11, etc.) out of this discussion, as converting one of those to a twin doesn’t involve any change in the number of lateral engines - just taking out the center tail engine and making the laterals somewhat bigger.  Also excluded are tail-mounted quadjets (VC10, Il-62), as these use conjoined engine pods (two engines per pylon), meaning that the two engines on each side could be swapped out for one bigger engine with little or no change in the stresses on the pylon and rear fuselage.
2: This doesn’t work for the A380, since not even the GE90 is powerful enough.  Sorry, A380!

Possible duplicate of Can the 747 be re-engined as a twin?

Yes, you'll get a wing redesign to allow for the higher weight and thrust of the engines: the inboard engine mounts now need to be twice as strong.
You can also run into ground clearance issues. An engine twice as powerful is often also larger. Here's a GE 9X engine being tested on a Boeing 747. You can see it's pretty close to the ground, and needs a really short pylon.

Going from 4 to 2 engines also means replacing half the instruments in the cockpit. You could save a lot of weight by removing the wiring for the outer engines, but that'd be a major operation.
Historically, the advent of high-bypass engines coincided with the introduction of widebody aircraft for a huge increase in passenger capacity and efficiency. So e.g. Boeing could make more money by offering a new design instead of selling 2-engined 707s.
For the remaining 4-engined airliners (B747 and A380) no engine is powerful enough to allow a 2-engine conversion. An engine powerful enough would be even bigger than the GE 9x shown above, and would scrape the ground when installed on a 747.
Finally, I'll leave you with a response to footnote 1: The VC-10 was used to trial the RB.211 engine, which was 2x as powerful as the original Conways:

3 different types of engine on that plane, assuming that tube under the wing is a small jet to offset some of the thrust of the tested engine.

The tube contains a set of electrical heating elements, to dissipate the power from the alternators on the RB.211. The balance between the RB.211 and the Conways was good enough not to need additional engines on the right side of the aircraft.

The thing you are proposing might be marginally feasible in podded-engines-on-the-tail, or perhaps the Comet, but there is absolutely no sense in converting a jet like the original 747 into a mega-777.

OK, so you've got 2 powerful engine instead of 4 weaker ones. So now the thrust is concentrated on 2 pylons instead of 4. This means completely redesigning the wing. Having 2 huge engines will also reduce wing bending relief. That probably means the flaps need to be redesigned too.
Moving the masses of the outboard engine into the inboard engine positions changes the CoG of the aircraft. So now the elevators need redesigning too.
In a single engine out situation, the sole remaining engine will provide almost the same amount of the original thrust on one side of the aircraft. And the aircraft must be able to either fly straight as if nothing happened or be in a long controlled descent. This means the to correct this yawing movement rudder needs redesigning too.

As you can see, this is in effect a brand new design. So why not start a new design and let it be a twinjet right from the start?
P.S.: This is why B-52 will never be a quadjet. Ever. Its rudder is just too small.


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What are the differences between C208 versions?
I am aware of a few versions of C208:

C208 Caravan
C208 Grand Caravan
C208 Grand Caravan Amphibian

Amphibian has pantoons with retractable landing gears, but what is the difference between C208 Caravan and C208 Grand Caravan? Is it just the cargo compartment in the belly? Are there other versions of C208? What are the differences between them?

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Change of Roll moment with change of wing position (z-direction)
If I compare two identical configurations, which only differ from each other in low wing and high wing position. We assume that the lift and drag remains constant. Will I get the same coefficients in moment for pitch, roll, and yaw? Because the force arms are the same?

I forgot to mention: i am looking at an axissymmetric (in x-direction) Body (tube) with a wing

For a first approximation, the coefficients will be similar enough, but not exactly identical. As you correctly point out the moment arms will be the same, but the interference effects of the wing-fuselage joint may not.
That is to say, your assumption that the lift distribution will remain unchanged will likely not hold out in reality.
The stability derivatives will not be similar, however.

If you have zero sideslip, then the moment coefficients will be similar between the two configurations. The only difference would come from difference in drag and its effect on moments, which would be small.
If you have non-zero sideslip, then the roll moment coefficients will be different due to dihedral effect.


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What is the main reason of air-cooled radials producing much less HP/L compared to liquid cooled Vs?
Is it because of the inherent limitations of cooling the cylinder walls, only 2 valves per cylinder, relative ease of scaling up instead of dialing up the boost, or some other reason?

The Merlin is .8 and the R2800 .89 lb/hp.  But does the Merlin's number include the coolant and radiator?

Radials do not produce "much less" HP/L. By the end of WW II, radials and inline engines were all very similar in horsepower to displacement numbers. Most had max horsepower numbers approaching up to 1 horsepower to 1 cubic inch.

I read somewhere once, maybe I can find it again, but it is because of the inherent limitations of cooling the cylinder walls. Air cooling is not as controllable, because the temperature of the air varies with altitude and season. Hence, the temperature of the cylinders in an air cooled engine are not as tightly controlled as a liquid cooled engine. So, the cylinders need to be able to operate over a greater temperature range. Since metal expands with temperature, that means the gap between the cylinders and the bore is typically greater in an air cooled engine than a liquid cooled one. So, this greater gap, even with piston rings means some of the cylinder head pressure after combustion escapes rather than act on the piston. Makes sense to me...

A liquid cooled engine doesn't necessarily expose all its cylinders to the airflow, which reduces drag in comparison to a radial engine. Also, water is a much better medium for distributing heat from an engine so it takes much more heat to have a detrimental effect on LC engines (aka temperature above operating levels).
Additionally, the design of a LC engine (inline or V) require the cylinders to be arranged in banks where one camshaft can be used to actuate valves in multiple cylinder heads (thus producing more HP).

On the other hand, rigidity is very important for a military aircraft. While radial engines can lose some cylinders and still be operational, LC ones will fail soon after getting coolant leakage.

Well, LC cooling doesn't require cylinders to be organized as banks they could still be seperate blocks - it rather enables it by removing the need to expose each cylinder to (sufficient) airflow.

The aspects you mention may affect power to weight and overall efficiency due to drag from the engine, but none of them have any effect to power to displacement ratio!


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Can a CIWS engage aerial targets?
Can a Phalanx-type Close-In Weapons System (CIWS) engage small aerial targets like drones, UAVs, rotorcraft? Is there any case that Phalanx was used against aerial targets?

What else is it supposed to do? It sure isn't anti personale or anti tank, right?

Considering that these systems are tested against drones, then the answer is yes. The limitations on what is too small to engage for any particular system are likely classified.

I'd love to know how they avoid shooting down every passing bird.

There might just be a human involved, @RobinBennett. Someone monitoring a video feed to ensure that it doesn't trigger on a flock of birds. Or, it may be up to de-noising algorithms to pick them out.

Or, maybe that’s how they feed the crew. JK. There is probably redundancy built into the system. For instance, the system is only active when underway, away from any friendly shoreline. Once active, the system can tell the difference between something natural and something man-made by its size, speed and pattern of flight.

It should be noted that the drones of the eighties were of a different size from what we think of as drones now.

It is designed to defend against incoming missiles.  A slow moving drone would be like shooting fish in a barrel.

I wonder if DJI Phantom (the only drone I know by name, so not advertizing here :) has enough RCS to be detected and targeted by Phalanx?

Phalanx can track and destroy missile size and speed targets, so as long as the drone or UAV has sufficient radar cross section, there would be no reason why Palanx would be unable to engage them.
The feasibility and utility of Phalanx, and similar systems against small drones and such is limited however. Imagine the risk of collateral damage if they were used in or near populated areas. There is a reason this is mainly a naval weapon...
At high seas, in turn, small drones, UAV's and rotorcrafts are not that much of a threat due to their limited range.


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Where do the forces exerted on flight yoke/stick comes from and how do they vary with flight conditions?
What are the forces felt on the flight yoke/stick based on the deflection of control surfaces and the aerodynamic forces acting on the control surfaces with respect to different aircraft attitudes?
For example: At low airspeed, the controls usually feel soft and sluggish, and the airplane responds slowly to control applications. At high speed, the controls feel firm and the response is more rapid.
Are there any equations that represent the same and how is the force exerted on the flight control yoke calculated?
My question is based on the control loading system of professional grade flight simulators for general aviation aircraft. I want to design a force feedback mechanism for a flight simulator to enhance realism.

This is very much application specific. For example in a 177 the elevator "feels" a lot different than a 172 at just about any speed. This is due to the design of the stabilator vs elevator. I don't think there is a generic "Xft-lbs per knot" if that is what you are looking for. Professional grade simulators only simulate one kind of aircraft, so they can base it off of specific aerodynamics.

Lets say I simulate this for only light category aircraft. Can I then generalize the aerodynamics then?

I've worked with control loading systems quite a bit. What you describe is a reversible system, where the flight crew feels the aerodynamic forces acting on the primary control surfaces (elevator, ailerons, rudder).

In the aircraft, the pilot pulls on the flying control, which is connected via a steel cable to the control surface, which then deflects. Airflow blows back on the deflected surface, with a force proportional to deflection of the surface. In the picture above $H_e$ is the hinge moment on the elevator, the aerodynamic part of which is $$H_{ea} = C_{he} \cdot ½ \cdot \rho \cdot {V_h}^2 \cdot S_e \cdot \bar{c_e}$$ with $C_{he}$ = moment coefficient, $\rho$ = air density, $V_h$ = airspeed at horizontal tail, $S_e$ = elevator area, $\bar{c_e}$ = mean aerodynamic chord of elevator.

The hinge moment coefficient is a function of trim tab deflection and local angle of attack (and therefore of elevator deflection), and is measured in flight tests. Figure below shows $C_{he}$ for the Fokker F27.
On ground, at airspeed zero, $H_{ea}$ is zero but there is contribution of friction, bob weight and down spring.

In the simulator, the same flight control is installed as in the aeroplane, but now is attached to an active actuator that is programmed to push back at the stick with the same force that would be experienced in the simulated flight state in the aeroplane. The actuator features an inner (control) loop and an outer (force feedback) loop.

Force vs. position and position vs. time measurements are taken from flight measurements, which must be tightly reproduced in the simulator in order for the sim to be certified. Reversible controls of small, simple aeroplanes are much harder to simulate than the irreversible, hydraulically actuated flight controls of larger planes, since these do not change as a function of flight state.

The pictures are from an old paper format TUDelft uni book on stability & control by prof. Gerlach

Have fun simulating friction!

First of all, thank you for such a beautiful explanation. I understand the concept of reversible flight controls. if you could also guide me as to how I familiarize myself with all such equations that exists and also how can I get access to the flight measurements part as well.

@JZYL We’ve dome so for numerous aircraft, at both the stick position and the surface position. Total simulated friction must match that of the measured aircraft, friction balance between fwd/aft appears during matching step responses and hydraulic system malfunctions.

@DhruvBhatt Studying aerospace engineering helps with the equations. The flight measurements are Intellectual Property of the company that mounted the necessary transducers on the plane and carried out the measurements during the course of a couple of weeks, in case of reversible controls. Irreversible control forces can be measured on ground with a hydraulics chart attached, using a dedicated force measurement set.

Can the flight measurements be extracted using a flight simulator software such as P3D or XPlane?

The flight measurements must be done with a real aircraft. They act as a baseline for setting up the flight simulator.

Similar to Elevator Hinge Moment, can you please explain the equations for Rudder and Aileron Hinge Moments as well?

Let us continue this discussion in chat.

Hi Koyovis, nice graphics, where did you get them from? The style somehow really reminds me of Aerospace Engineering at Delft University. Can you add a reference?

@DeltaLima Indeed, you've recognised the source of the images. They are mentioned in the image link, visible when "edit" is clicked. The book is old but timeless.


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When flying commercial a/c form US to Latin America by GPS which agency controls the flight?
When flying internationally from the US to Latin America, the use of GPS is common.  What agency or agencies control the flight and how?

Over the ocean, most flights aren't under control at all. Over land, whichever country you are over uses their ATC.

What does GPS have to do with who controls the flight?

@zeta-band Well, I hope the pilot is in control. :)

Welcome to aviation.SE! I'm not sure what you're asking and you might find the [tour] is useful to understand how this site works. If you can tell us more about what "controlling the flight" means then you may get a better answer, e.g. are you asking which Air Traffic Control units direct the flight if GPS is not available?

Airline flights are almost ALWAYS under ATC control. The entire world is divided into Flight Information Regions. ATC control is passed from one agency to another as the aircraft passes each FIR boundary.
It is not a requirement for an aircraft to have GPS equipment as there are other navigation systems which can be used for navigating airways or direct routings.
Here is an interactive map that shows each FIR and it's boundaries: ICAO FIR WORLD Map

A flight from Miami to Bogota would fly though 5 different Flight Information Regions.

I did a trip from S America to Canada once in a jet and the biggest challenge was understanding the controllers.  Their equipment is old so the audio sucks, scratchy and muffled, they use different phraseology, then on top of that, the Spanish accents.  The airports we stopped at didn't have ATIS so they give you all the ATIS information buried in your arrival clearance.  Quite a challenge on a first time/only time trip.  First time talking to a US center controller on the last leg, it was like listening to a DJ on FM radio.

Note that airline flights are rarely in a straight line (even in this case where great circles don't really matter) due to weather, overflight fees, political disputes, etc.  So, the proposed flight could end up going through a slightly different set of FIRs than shown.


text:
Is there any book like the "Stick and Rudder" about the helicopter?
I have read the Stick and Rudder, that's really a classic book. Now I want to study the flight theory about helicopters, is there any book like this?

Have you read the FAA's Helicopter Flying Handbook?

I downloaded it, but it isn't the kind of book I wanna.I need a simply and humorous book without too many equalitions

There's a book called "Cyclic and Collective" by Shawn Coyle. It's a good book, although it is a little bit more detailed than something like "Stick and Rudder". There's a lot of good stuff in there, though.


text:
Can I derive the wind speed and direction?
If I know the IAS and GS (with directional components) of an aircraft, can I derive the wind speed and direction?
I'm working on an application that simulates the flight of an aircraft.  Another application that takes as input, the computed trajectory of my aircraft, which includes the IAS and GS of the aircraft (at points along the trajectory) needs to know the wind speed and direction at those points.
I think it is possible, but am not sure, to isolate the wind speed and direction using the IAS/GS (or both).
Can someone confirm this?  Is this a reliable way to do this?  Do I need other information to complete the calculation?

Possible duplicate: How is the wind direction and wind speed measured in flight?

Yes, in part; I had read that but still needed to ask for my own clarity.  I just needed to confirm for myself that it was possible to extract the wind at a point from the data available (in my case, I have IAS, GS and TAS, thought in my question I thought I needed IAS and GS only, but the answer cleared that up).

No, you need the True AirSpeed (TAS).
The Indicated AirSpeed is not a realiable measure of it, since it is not even calibrated against the instrument mounting position and is not intended to measure the speed relative to the airmass (what you actually need to derive the wind components).

I use to fly a small gyro. It's fitted with an old, simple GPS, so that I always know the instantaneous ground speed.
If, while flying level and at constant engine rpms, I describe a complete turn, there's a point of that circle where the groundspeed reaches a minimum. The gyro's heading is, at that point, the wind direction. And the wind speed is the difference between the airspeed (TAS) of the anemometer and the ground speed as shown by the GPS...


text:
How come the hover efficiency for tiltrotors is better than tiltwings?
How do you compare the hover efficiency of tilt-rotor and tilt-wing aircraft? The Wikipedia page Tiltrotor mentions:

Tiltrotors generally have better hover efficiency than tiltwings (...)

But I was expecting just the opposite because of the downward thrust loss due to wing blockage on tilt-rotors.

The tilt rotor concept power requirement in vertical mode is 1.5 times higher than the same payload capabilty helicopters and in horizontal cruise 3 times higher than the same speed propeller aircraft.
The V-280 Valor is able to carrying 14 soldiers or lifting up 4500kg payload externally, but it requires 2x5000 shaft horsepower.
To decrease the blade loading in vertical mode and increase the propeller loading in aircraft mode, the Bell already studied the variable diameter propeller, but I assume, the ultimate solution would be the variable swept scimitar design propeller blades. This design is continously adapting to the large diameter, slow speed, low disc loading helicopter regime and to the high RPM, low diameter, high disc loading aircraft mode. The pivot point would be offset from the propeller blade axis, so the diameter is decreasing by the rotation. Of course the pitch and cyclic control remaines the same.


text:
What limits the speed of a hypersonic aircraft flying in the lower atmosphere? Are there estimates for its max possible speed?
I'm guessing it is the heat resistance of the missiles leading edge material.
Hypersonic cruise missiles all seem to attain Mach 6 and below. The boost-glide weapons fly faster but up much higher. I think the Russians have ABMs that make it to Mach 10, but that is also intended for high altitude intercept. The X-15 attained about Mach 6, again up around 100,000 feet.

You might need to factor in the power available vs the power needed to overcome air resistance.  Consider why rockets start straight up, and gradually bend their trajectory to the horizontal as they get above atmosphere.

The limit is imposed by the maximum dynamic pressure max-q the aircraft can sustain. Rockets are usually slowed down during their ascent, before reaching the max-q point, in order to stay within structural limits.

Consider that a Saturn V burns all its 1st stage fuel in about 2.5 minutes in order to accelerate to about 6,000 mph (at 42 miles altitude).  A bit of searching suggests only about 1/5 of a rocket's energy is used to increase altitude, the rest is to accelerate to orbital velocity: https://space.stackexchange.com/questions/35858/how-much-of-a-rockets-energy-is-used-to-achieve-altitude-vs-horizontal-velocity  So achieving sustained high speeds in level flight, especially at lower altitudes, takes a heck of a lot of fuel.

@jamesqf That's comparing apples with oranges. The main reason that the early stages burns so much fuel is because it's accelerating the rest of the stages, not the low altitude. Orbital velocity is Mach 23 at sea level...

@Sanchises: So have you got better figures for fuel consumption at high speed & low altitude?  AFAIK rocket launches are the only practical example that comes close - at least that's not classified secret :-)

Another example is Project Pluto, a nuclear ramjet powered cruise missile designed to cruise indefinitely at low altitudes at Mach 3.


text:
What is the difference between aircraft integration and assembly?
I have never had a clear understanding of the difference between aircraft assembly and integration. My understanding is that putting together the airframe alone is assembly and putting together the avionics and other systems in order to get the aircraft flight-ready is integration. I'm not sure if this alone captures the difference. It would be great to have inputs regarding this.
Thanks in advance!

Welcome to Av.SE!

integration is the process to figure out what/how to put together, while assembly is actually putting things together. one is done by a team of engineers, while the other is done by a factory of workers.

It makes sense. So Integration comes first and then assembly. Thanks.

No. The process of  figuring out how to assemble something like an airframe is called Process Planning, or Methods. Your question post is closer to the answer.

I was many years with an OEM.  Integration is process of coordinating and incorporating, harmonizing you might say, the unique engineering of systems and processes of two engineering and production organizations or products, to make them work together in the final product.  To integrate them.
For example, a manufacturer designs and builds an airframe.  An avionics suite, or some other system, designed and manufactured by a second party, not specifically for a given type, is selected to be used in this new airframe.
Someone has to integrate, or adapt, the pre-existing avionics suite into the airframe and systems of the host.  The integrator is responsible for all of the activities of incorporating that system;  where/how it fits, how it interfaces with host systems, etc. In other words, the integrator is responsible for making the second party system work smoothly in the host system.


text:
General Rule for Calculating Center of Gravity of any Aircraft Configuration
I'm looking for method(s) to determine the CG of any fixed wing aircraft for it to be reasonably stable. I only know for conventional aircraft, placing the CG between 1/4th and 1/3rd of MAC behind MAC leading edge usually works but when it comes to unconventional configurations like canards or tandem wings or even more lifting surfaces I don't know what to do.
I drew some examples to ask what kind of approach should I use for each of these aircraft configs to find their desirable CG position.
Also how can I decide if a surface is better suitable as a positive or negative lift producing surface? For example for a canard we know that both wings produces +lift but if we shift the CG on the forward wing as if it was a conventional airplane the main wing becomes a down force generating surface (assuming it can fly) doing that is obviously wrong but when it comes to model B or E it gets blurry for me.

A and C have some merit.  Throw the rest away. Center of Gravity needs to be placed first where the plane is directionally stable and directly under the net lift vector of the entire aircraft.  Next determine what pitch characteristics you want.  To be staticly stable, the aircraft must pitch down below its trim speed and pitch up above it. The only difference with a canard is that it provides upforce from the front rather than downforce from the back.
In this sense canarded or tailed aircraft are actually bi-planes.  The wing in front must stall first to be safe.

"Throw the rest away." I disagree. I would love a chance to test-fly type E. - I want to help decide where the engine should go.

Instead of E, what ever happened to the one that swept the entire wing as one?


text:
What if the bomb bay is hit?
The bomb bay on s heavy bomber is a somewhat big target. If the ordnance were to explode, it would be very dangerous for nearby aircraft. But I haven't really seen much evidence of concern over this during the massed air raids in WWII, Korea, or Vietnam. What am I missing?

No heavy bomber air crew was ever known to register a complaint about the sensitivity of their ordinance to enemy fire.

Are you asking about the situation where flak or enemy aircraft rounds penetrate the bomb bay doors and strike a bomb? I would imagine (but my search-fu isn't helping me confirm) that the bombs weren't actually armed until they fell out of their bomb racks, therefore they wouldn't be terribly likely to explode within the aircraft.

I definitely have read stories of WW2 bombers exploding into nothingness because the entire bomb load detonated under enemy fire.  But I don't think it would be practical or even useful to try to armor the bomb bay against a direct hit from a large-caliber AA shell.  With that kind of a direct hit the airplane's going down regardless of whether the bombs explode or not.  Is there some smaller shell caliber against which it would make sense to try to provide armored protection for the bomb bay?  Hmm, dunno--

Reminds me of the story of plane reinforcements being designed on the basis of damage seen when they came back from battle.  The problem being that the places needing reinforcement were on the planes that didn't come back.

Major Kong will just have to go down to the bomb bay and fix the problem   https://www.youtube.com/watch?v=is5u5uhyc6s

Modern explosives are remarkably unwilling to go off without the correct primer. Some of them will not explode even when set on fire (and just burn in that case).

https://aviation.stackexchange.com/a/78764/31425

High explosive is relatively less sensitive and requires a primer's blast wave or high-speed impact from a metallic object (speed ≥ 1000 m/s) to detonate.

https://link.springer.com/chapter/10.1007/978-981-19-2856-7_2
The muzzle velocity of the "88" antiaircraft gun, which was common enough to be the origin of the word flak, was only 840 m/s.
Its shell was about 9 kg, thus with a kinetic energy of about 3.2 MJ.  Even at a fairly low cruising altitude for a bomber of 15000 ft or 4500 m, the shell would have converted about 0.4 MJ of that to potential energy, leaving it with a speed of about 790 m/s.
Although a high caliber bullet from an airborne attacker might reach 1000 m/s, it will have slowed down after penetrating everything between the airplane's skin and the 'meat' of a bomb.

Edit, from comments:
BF 110's, when attacking bombers from below with ammunition (often just 30 mm, and not big AA shells) could detonate bombs.  I don't have authoritative confirmation of such detonations, but Wikipedia says that

The optimum target for the night fighter was the wing fuel tanks, not the fuselage or bomb bay, because of the risk that exploding bombs would damage the attacker.

So, at least for aerial attackers, the reason that bombs still on board were rarely reported to have been exploded is that the attackers deliberately avoided that.

Weren't the AA shells fired by the 88 explosive?  After the shell explodes, some of its pieces could be going pretty fast.

@WayneConrad even without solid lumps moving at speed, sympathetic detonation can be caused by shock alone, even in some compositions that won't cook off

The threat is sympathetic detonation, not the charge being hit by a material object hard enough to detonate it.


text:
Combating flutter with mass distribution?
Considering aeroelastic flutter, when an eigenfrequency of the wing meets with an aerodynamic frequency, the commonly mentioned solutions make sense: avoid the convergence zone by limiting speed, or by increasing the wing's eigenfrequency trough stiffness, reduced weight, or reduced span.
Makes sense, especially if I use the spring and mass model for the wing:

I'll use the analogy with the conventional spring as it makes things easier to understand for me.
So Wikipedia says: "Prediction involves making a mathematical model of the aircraft as a series of masses connected by springs and dampers which are tuned to represent the dynamic characteristics of the aircraft structure. [...] Small carefully chosen changes to mass distribution and local structural stiffness can be very effective in solving aeroelastic problems."
My 3 questions are:
1: Do I understand correctly that with this approach, the wing will act similarly to the following model, where the motion is rather chaotic?

2: This should have no certain frequency. Does this mean flutter is avoided?
3: Does this mean flutter can be avoided by redistributing mass around the wing, without increasing stiffness or reducing mass?

so flutter is avoided here by having so many random eigenfrequencies in a single object that neither can really get excited much?

According to my understanding, yes.

3: redistributing mass without reducing mass means increasing mass or introducing weak spots.  That might avoid flutter, but it would definitely harm performance in many other ways.

Do I understand correctly that with this approach, the wing will act similarly to the following model, where the motion is rather chaotic?

No, your combination of masses and springs will have several eigenfrequencies. For aeroelastic analysis only the lowest eigenfrequency counts because this will couple with aerodynamic frequencies at the lowest flight speed. Once you fly faster, the next eigenfrequnecy in line will resonate, but that is only hypothetical because flutter caused by the lowest frequency has already destroyed the airframe.
In very rare cases the lowest eigenfrequency will already couple with aerodynamic oscillations at speeds below stall speed. Then the aircraft can be operated in the gap between the lowest and the second-lowest eigenfrequency.

This should have no certain frequency. Does this mean flutter is avoided?

No, since it has several eigenfrequencies, flutter is as certain as before.

Does this mean flutter can be avoided by redistributing mass around the wing, without increasing stiffness or reducing mass?

Flutter cannot be avoided, only its speed shifted up by damping. The rule always is to place the center of gravity ahead of the axis of rotation (that would be the elastic axis in case of wing torsion), so any movement causes an inertial damping force. This is the reason for the forward location of wing mounted engines.

You could mention that hydraulic control surfaces operate unbalanced, with the CG well aft of the rotation axis by virtue of the "rigid" hydraulic connection that makes the surface effectively a fixed part of the trailing edge, and if unpowered, the damping ability of the idle actuator is considered sufficient to prevent the imbalanced surface from reacting/magnifying movement in parent structure.  If the damping ability of a dead actuator(s) is found to be insufficient, then additional passive flutter dampers are required.

"The damping ability of the idle actuator is considered sufficient..."  VSS Enterprise?  For moving/turning surfaces, how stable or unstable to be in the airstream seems to be the quandary, especially when Velocity goes well beyond hurricane force.


text:
What amount of lift would be created in the fallacious "lift from equal transit time" theory?
Context
A wrong --but simple to understand-- theory for lift states:

Air molecules that were close before being separated at the leading edge must be close again after they join back at the trailing edge. For this to happen, molecules must travel the two paths in the same time.

As the path is longer on the upper side, due to the camber of the wing, air is accelerated on the upper side. By Bernoulli's equation, static pressure decreases when velocity increases, so the pressure is lower on the upper side, and higher on the lower side.
Lift is the force created by this pressure differential on the wing sides.

(What is wrong is not Bernoulli's equation, but the assumed relationship between path length and speed.)
Such theory is, for instance, presented as correct in this book.
More at Nasa.

Question
We know this theory is wrong and lift created by path length difference would be a very small percentage of the lift actually measured (and even null for symmetrical airfoils).
Let's assume a rectangular wing with a simple cambered airfoil, at the critical angle of attack:

What would be the pressure differential according to equal transit time difference? What percentage of actual lift does it represent?

Excellent question, though I'm afraid I'm not going to understand a word of the math that this will turn up. :( Someone (ahemPeterKampfcough) will come up with a final number and I'll understand that, though.

That book is an absolute joy to read. I've never seen so many fallacies condensed in so little space. As it turns out, induced drag decreases lift, flaps are used to trim the aircraft, there are slots [sic] in front of the wings to reduce the air 'piling up', airliners are equipped with a steady supply of oxygen and the cabine is only pressurised to keep your blood from boiling.

Notice that, if split parcels rejoin, then the spanwise circulation is exactly zero, regardless of airfoil shape.   Zero circulation gives zero lift.  "Transit-time theory" is actually a recipe for producing zero lift.  Or said differently: spanwise circulation is permanently splitting the upper/lower parcels, and the larger the split-distance, the greater the circulation, and the higher the lift force.

The theoretical lift coefficient of a NACA 4412 at the critical angle of attack (about 16 degrees) is approximately 1.7.
\begin{equation}
c_\mathrm{l}=1.7
\end{equation}
For our purposes, that same airfoil has its upper "surface" about 1.03 times as long as its lower, which can be seen by using the distance formula on the coordinates or integrating the cambered NACA airfoil equation. Here is the distance-formula method applied to the Airfoil Tools numbers (thanks to ROIMaison for making me check this calculation):

On average, therefore, the fallacy would predict the air moving along the upper surface at about 1.03 times the velocity as along the lower in order to join up properly.
Making all the assumptions from the NASA link, Bernoulli's equation says
\begin{equation}
P_1+\frac{\rho V_1^2}{2}=P_2+\frac{\rho V_2^2}{2}
\end{equation}
Let's represent the lower surface with subscript 1 and the upper with subscript 2. The respective velocities are averages across the surfaces. We have
\begin{align}
P_1-P_2&=\frac{\rho (1.03V_1)^2}{2}-\frac{\rho V_1^2}{2}\\
&=\frac{609}{20000}\rho V_1^2\\
\end{align}
So the average difference in pressure coefficient (i.e., the lift coefficient) is given by
\begin{equation}
\mathrm{avg}(c_\mathrm{p_{ETT}})=\frac{609}{10000}\frac{V_1^2}{V^2}=c_\mathrm{l_{ETT}}
\end{equation}
where $V$ is the freestream velocity and ETT stands for equal transit time.
We have another expression for pressure coefficient that we can apply to get a final value. We end up with
\begin{equation}
c_\mathrm{l_{ETT}}=\frac{609}{10000}\frac{V_1^2}{V^2}=1-\frac{V_1^2}{V^2}
\end{equation}
or
\begin{equation}
\frac{V_1^2}{V^2}=\frac{10000}{10609}
\end{equation}
and therefore
\begin{equation}
c_\mathrm{l_{ETT}}=\frac{609}{10609}\approx.0574 \approx .034c_\mathrm{l}
\end{equation}
which is less than 5% of the actual value. However, it is important to keep in mind that the fallacy predicts no lift change with angle of attack (just one of the many reasons it is silly), and so the relationship between $c_\mathrm{l_{ETT}}$ and $c_\mathrm{l}$ varies with angle of attack for a given airfoil. For the 4412, they meet at just over a -5° AOA. For a symmetric airfoil, they meet at zero AOA. For any other airfoil, the calculation must be redone with the correct path lengths and actual lift coefficients, although magnitude of the result is not likely to change dramatically.

Do you have any evidence/support for "that same airfoil has its upper "surface" about twice as long as its lower (this can be seen by using the distance formula on the coordinates or integrating the cambered NACA airfoil equation)"? It seems like a lot to me

I had an extra term in the distance calculation that made the result way too much. Good catch! I've edited the result (a substantial change) and I appreciate your pointing this out.

That's an excellent answer, the kind I was looking for.

Your calculations underestimate the lift predicted by the "equal transit theory", because you measure upper and lower surface length from the nose rather than from the stagnation point. For airfoils with a large nose radius, this difference is substantial. When surface length is measured from stagnation point to trailing edge, this theory not only predicts lift that changes with angle of attack, it also predicts lift for symmetric airfoils. It is still wrong, though.

@RainerP. Fair enough. When I was taught ETT (and yes, I was taught ETT in an actual school, albeit a primary school) nothing so physical as a stagnation point was even hinted at. I guess to make a complete comparison one needs to first rigorously establish how incorrect the baseline incorrect "theory" is...

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Is it legal to fly over Vatican City?
In February 1922 one or more aircraft flew over the Vatican low enough to disturb the papal conclave then underway. In response, authorities forbade flights over all of Rome until the conclave was concluded.
About seven years later, the Lateran Treaty between the Holy See and the King of Italy included a provision forbidding flights over Vatican City.
Various changes both in technology and in regulation of aviation have happened since that time. Ninety-one years after the aforementioned incident, Pope Benedict XVI announced that he would resign as of the last day of the month, and on February 28 he boarded a helicopter in Vatican City to fly to Castel Gandolfo.
What are the current rules and practices on flights over Vatican City?

SkyVector shows a prohibited area LIP-212 over Vatican, but for most prohibited areas there is still a way to get a permit, presumably in this case on approval by Vatican administration (details should be in the Italian AIP).

Pope is also special (like presidents and prime ministers): special rules and procedures (between civilian and military). If a state authorize a flight, it should guarantee safety (according one of the Vienna Conventions, IIRC), and other stuffs (for such reasons, sometime head of a country are not allowed to flight other country, until a full security protocol is established)

I think the reference to the Pope flying in a helicopter into and over Vatican City is a non sequitur. There are plenty of pilots that fly into and through prohibited airspace. Presidents, popes, and extra-terrestrials (including astronauts) to name a few. Marine One flies quite often into prohibited airspace. It is a given. Look at who is prohibiting flight in the area and for what purpose? They are the authority that can grant exceptions in most cases.

@JanHudec Your conclusion is correct, but the restricted area covering the Vatican City is P243, not P212

Yes, it is legal to fly above Vatican City, but not below 2500 FT if flying IFR or 3500 FT if flying VFR.
There are several Prohibited Areas established over Rome. Prohibited area 212, mentioned by @JanHudec, is not actually above Vatican City, instead it is there to protect the Regina caeli prison.
However, there is a prohibited area covering the Vatican City: P243, in which IFR traffic is prohibited below 2500 FT and VFR traffic is prohibited below 3500 FT. The AIP (ENR 5.1.1-14) notes that the area is "subject to overflyng restriction for security-safety reasons". While not mentioned directly in the AIP, presumably, there is a way to get permission to fly into the area in special cases, for example to access the Vatican City Heliport.

The red area in the image above shows prohibited area 243. Marked with blue is the heliport.
There appears to be no restrictions for traffic above 3500 FT. There is even an airway (L5) almost directly above the Vatican City (black line on the map), which has a lower limit of FL95 (9500 FT) - which is a completely normal lower limit of an airway, Vatican City or not - so it seems IFR traffic will routinely fly directly above, albeit not a low levels.


text:
Why does adverse yaw exist?
Why does the aileron deflected upwards, thereby reducing lift, not also reduce drag on that side?

One significant component of adverse is yaw is due to roll rate not aileron deflection-- there is sometimes a tendency for the nose to yaw toward the rising wingtip, even if the roll is caused by spoilerons rather than ailerons.   However the adverse yaw component due to aileron deflection can also be very significant, and can be seen in some demonstrations that actually involve zero roll rate.  As to why-- well...

The wing that goes up  trades kinetic energy for potential energy and the other wing does the opposite.

Indeed, when an aileron moves upward, it locally generates less lift and less drag.
Assume we are talking about the aileron on the right wing.
The reduced lift drops that wing, rolling the aircraft right. But the reduced drag accelerates that wing, yawing the aircraft left.
On the left side the opposite happens, but it has the same effect. The aileron moves down, increasing lift to roll the aircraft right. But the added lift increases drag, decelerating the left wing to yaw the aircraft left.
That is the adverse yaw effect.

When the left aileron goes down it creates extra lift, and drag, but the right aileron travels up and also creates drag. Why is there more drag on the left aileron than the right one?

@Zebrafish the one going up reduces lift, so it reduces drag. Only when it moves up so much that it  produces negative lift the drag will increase again, but still less drag than the opposite side

The aileron going up does produce drag as well - once it is producing a sizeable amount of downforce. At rest, it is aligned with the back end of the wing, and in a wing with no camber at zero Angle of Attack the aileron produces symmetrical drag. But wings usually do have camber and/or AoA ≄ 0, so at rest the aileron points trailing edge downwards and starts with positive lift plus the associated drag. When deflected upwards, lift and drag first reduce until the aileron is flush with the airstream, while the other aileron produces more lift and drag.

Adverse yaw has two causes:

The "hooking" effect described above, and depicted in the figure from my old uni book. It shows the other way to explain the adverse yaw drag: the wing producing more lift also produces more induced drag.
The roll rate resulting from aileron deflection, which increases Angle of Attack of the downwards travelling wing. This tilts the local lift vectors of rising vs. falling wing, and the horizontal component produces an additional adverse yaw effect. As depicted below: $\bar{\Delta c}_rn$ points upwards at one wing, and downwards at the opposite wing, at different tilt..

Where is the lower picture from? It's kind of confusing... or I'm just tired...

And regardless of who's confused, you should credit the original author - even if it is yourself.

The pictures are from a university course lecture handout by prof. Gerlach, released in 1981 and as far as I know still used today. Not available on the open market and in Dutch. As you can see, I’ve held on to it and still use it as a reference.

Now I see where English translations have been pasted onto the diagram.

I've read about 6 explanations of adverse yaw and my reaction each time was "Yeah the upward travelling aileron creates drag as well". I feel all the explanations are just handwaving it away with a stock explanation. Though I don't if this explanation is correct, it definitely makes sense to me unlike the others I've read.

The principal drag in question is induced drag, that is the drag necessarily induced by creating lift.
The aileron that goes down increases lift on that wing, and thus also induces more induced drag; conversely, the aileron that goes up reduces lift on that wing and thus induces less induced drag. Hence, the aircraft will tend to yaw towards the aileron that goes down.

So once the roll rate becomes constant rather than increasing-- implying that both wings must be making the same amount of lift-- the adverse yaw goes away?  This answer would seem to suggest so, but we don't see this to be the case in reality.

@quietflyer If the roll rate is constant how do both wings make the same amount of lift? I think your point is right though, that once the roll rate is constant the major cause of adverse yaw is the other explanation. It's the rotation of the actual wing that causes it, not the aileron deflection itself.


text:
Calculate parasite drag coefficient from drag polar
The drag polar of an electric-powered unmanned aircraft is
$$ C_D = 0.038 + 0.0458 C_L^2 $$
What is the parasite drag coefficient?

Well, $C_D = C_{D,0} + k_1 C_L^2 + k_2 C_L$ according to Brandt et al's "Introduction to Aeronautics: A Design Perspective," so the parasite drag coefficient is $C_{D,0}$ or 0.038.
(Also, $k_1 = 0.0458$ and $k_2 = 0$.)


text:
Why isn't Yamaha EPeX300Ti more popular in aviation?
The answer to this question might be obvious to experienced aviators. But as an armchair enthusiast, I really couldn't find a satisfactory answer anywhere on the internet.
Below two pages came from companies that sells this engine
http://www.paramotoraviation.com/products.php?product=EPeX300Ti-%252d-300HP-Aero-engine
https://www.universalmarinesurveyingcenter.com/product/yamaha-epex300ti-300hp-aero-engine/
Below is an article talking about one successful application of this engine
https://www.kitplanes.com/fly-it-like-you-stol-it/
On paper, this engine outperforms other common small piston aviation engines.

300 hp
4 cylinder
turbocharged
similar weight to Rotax 915iS

Fast planes such as Lanciar Legacy and Glasair III have engines output about 300 hp but are much heavier. Rotax 915iS, one of the most advanced small aviation engine, output about 145 hp.
Looking at the performance alone, Yamaha EPeX300Ti should dominate the market.
Why is there so little attention to this engine? Safety? Design flaws?

Steve Henry flies one.   That’s a feather in their cap.

@jim yes, one of the links talks about Steve Henry's airplane. That's how I learned about this engine.

If your car engine dies you coast to the side of the road, if your airplane engine dies chances are you do too. Power to weight ratio is only one factor in choice. Also keep in mind it's a conversion not a purpose built aero engine, so support is a big concern.

This question is too broad and opinion based. The answer will be different for every owner. The super short answer is individual value.
But value is subjective and wraps up capital investment, reliability, maintenance, legal certification hassles, frame conversion hassles, confidence in finding parts and mechanics for the engine in 10 or 20 years, trust in the company, fuel burn, resale market value, even minor stuff like noise issues or shipping cost.


text:
Historical Delta Airlines Tail numbers (2012) - Where to find?
I was looking for the tail numbers of 2 flights.  The only records I could find were the US Transportation data for domestic flights but I could not find any International Delta Airlines Flights.
The flights are:
Delta DL19 from London Heathrow to Detroit Wayne county on 15th June 2012 departing 09:02 landed at 12:23
Delta DL46 from Atlanta Hartsfield Jackson to London Heathrow on 25th June 2012 departing 17:59 landed at 07:36 (on 26th)
Which databases can I look at?

This question belongs to the passenger area of aviation, you should ask at Travel.SE.

FlightAware likely has the data, but free access only gives you the past three months of a given flight number.  You'll need paid access to get at the full history of a flight.  FlightRadar24 is another site with similar capabilities.


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Why does Vx have a relationship with thrust but Vy has a relationship with power?
https://www.boldmethod.com/learn-to-fly/performance/vx-vy-altitude-and-where-they-meet/
After I watch this article, now I understand the relationship between altitude and the rate of climbs.
But I wonder why Vx has to do with thrust and Vy has to do with power.
Could you explain them for me?

For the best climb speed, you need to maximize your vertical speed $\text{v}_z$.
For the best climb angle, you need to maximize the ratio $\frac{\text{v}_z}{\text{v}}$ of vertical speed $\text{v}_z$ and flight speed v.
And, as Robert correctly notes, thrust is power divided by speed. $\text{T}=\frac{\text{P}}{\text{v}}$. Do you see a parallel?
Please see here for what needs to be considered to calculate climb speeds at various altitudes. To simplify things, we can model the airplane by using the quadratic drag polar and assume that thrust depends on air density and on flight speed to the power of a constant exponent n$_v$ which normally is -1 for piston airplanes. This gives us three equations to model lift L, drag D and thrust T. Weight is given and considered constant by neglecting fuel consumption.
$$L = m \cdot g$$
$$D = \frac{\rho}{2}\cdot v^2 \cdot \left(c_{D0} + \frac{c_L^2}{S\cdot AR\cdot\pi}\right)$$
$$T = T_0 \cdot \rho \cdot v^{n_v}$$
How thrust changes over altitude is explained here (besides much else). The equation is valid for a wide subsonic speed range but does not cover static thrust. Now we can express climb speed as the excess power left over after subtracting drag from thrust, times flight speed:
$$v_z = v\cdot\frac{T-D}{m\cdot g} = \frac{v}{m\cdot g}\cdot \left(T_0 \cdot \rho \cdot v^{n_v} - \frac{\rho}{2}\cdot v^2 \cdot \left(c_{D0} + \frac{c_L^2}{S\cdot AR\cdot\pi}\right)\right)$$
When we now express the lift coefficient as a factor of speed, the equation is ready for derivation with respect to speed. We do the same for the equation for the climb angle. Set both results to zero to find the conditions for the highest climb speed and climb angle.
$$c_L = \frac{m\cdot g}{\frac{\rho}{2}\cdot S\cdot v^2}$$
$$\Rightarrow\:v_z = \frac{v\cdot T_0 \cdot \rho \cdot v^{n_v}}{m\cdot g} - \frac{\rho\cdot S\cdot v^3 \cdot c_{D0}}{2\cdot m\cdot g} - \frac{2\cdot m\cdot g}{\rho\cdot S\cdot v\cdot AR\cdot\pi}$$
and
$$\frac{v_z}{v} = \gamma = \frac{T_0 \cdot \rho \cdot v^{n_v}}{m\cdot g} - \frac{\rho\cdot S\cdot v^2 \cdot c_{D0}}{2\cdot m\cdot g} - \frac{2\cdot m\cdot g}{\rho\cdot S\cdot v^2\cdot AR\cdot\pi}$$
Or you put them in a spreadsheet and plot them over speed. Either way. But what you can see already is that in the climb speed equation we have power (thrust times speed) on the right side whereas in the climb angle equation only thrust is left on the right side.
Having read the article you link to, I am not very happy with the explanations given there. The power curve is supposed to change as you climb - Nonsense! What changes is your true air speed, but the plot sticks with the same true air speed range as altitude increases. All it shows are ever smaller sections of the power curve, which creates the illusion of change where no change is. Plot it over IAS and see for yourself.
Nomenclature:
$c_L \:\:\:$ lift coefficient
$c_{D0} \:\:$  zero-lift drag coefficient
$n_v \:\:\:$  thrust exponent, as in $T \sim v^{n_v} $
$D \:\:\:\:$  drag
$S \:\:\:\:$  wing area
$T \:\:\:\:$  thrust
$T_0 \:\:\:$  density-specific thrust at a given speed and altitude
$m \;\:\:\:$  mass
$g \:\:\:\:\:$  gravitational acceleration
$v \:\:\:\:\:$  flight speed (as true air speed)
$\gamma \:\:\:\:\:$  flight path angle
$\rho \:\:\:\:\:$  air density
$\pi \:\:\:\:\:$  3.14159$\dots$
$AR \:\:$  aspect ratio of the wing

The angle a plane can climb at is entirely dependent on excess thrust.  With props, maximum thrust can be at a lower airspeed than Vbg, so even if the plane is not at its optimal wing AOA (more drag), the maximum line between drag and thrust is where you get your best climb angle, Vx.
That's the easy part.  How can a plane climb faster when it is climbing at a lower angle?  By going faster!
The formula for Power is Force (thrust) x Velocity
When you see this graph, note that the maximum distance between Power and Drag is Vy.  This plane will fly higher and farther for a given amount of fuel than at Vx.

Vx is conceptually pretty simple, a climb is very similar to a car going uphill. The steeper the hill the more excess thrust is needed regardless of speed. The only real difference is that a car only has parasite drag and can down-shift to increase thrust at lower speeds, so assuming infinite traction the VX of a car approaches 0 speed. In a plane there is a need to maintain lift equal to aircraft weight. The resulting induced drag and the efficiency of the prop at different airspeeds(The prop also has its own angle of attack) all play together so that maximum excess thrust happens at a particular speed; usually close to minimum total drag depending on the prop selection and engine torque curve.
Now Vy is conceptually a little less clean and doesn't have the direct car analogy(a car's Vy is also its Vx). But it is still a matter of going uphill. If the prop were optimized only for rate of climb then Vy would occur very close to but never below the point of maximum lift/drag ratio for basically the same reasons that best glide distance(angle) is at best L/D and best cruise range is always above best L/D.(Sorry, To be continued something just came up.)


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Which single pilot aircraft had a range of at least 2160nm in the late 90s?
I'm writing a book and looking for an airplane that can handle that distance (2160nm), was available back in the late 90s, and can be operated by a single pilot.
Thanks!

What is possible with GA airplanes is also certainly possible with private jets. A simple modification is all it takes.

@CGCampbell The Cessna Citation X has a minimum flight crew of two pilots.

Cessna Conquest II.

Mitsubishi MU2.

@acpilot The Cessna Conquest II looks good, thanks! The MU2 has half of the range, or is there a long range version?

Smaller planes that had this range back in the 90's were not so rare, but they were jets, most of them were not intended to be operated, at least commercially by a single crewmember. It would have been totally possible though.
When considering that you are writing a book, you may take some liberties (in addition to the crew count) when it comes to range: the official figures take into account the fuel reserve regulations, so after the declared range is flown, the plane would most likely be able to continue flight for about an hour before fuel starvation.
So, at least these had the range of 2160nm minimum (some not fully loaded):

Learjet 60
IAI Westwind
Dassault Falcon 20 (some variants)
Hawker 800

(To keep things simple, I'll turn this into community wiki right away)

It is intended as a private flight with passengers. How can you determine if a jet is certified for single-pilot use?

@root66 You can find that in the TCDS (Type Certificate Data Sheet). If you Google for TCDS and the aircraft type, they should be easy to find. E.g. for the Learjet 60, the EASA TCDS says: "Minimum Flight Crew: For all flights, 2 persons (pilot and copilot)", so it is not single pilot certified.

@bianfable Thanks! The complete Hawker series also has 2 pilots minimum. I haven't found the other two TCDS's yet.
Now, I could investigate further, or portray the pilot as careless for flying without a co-pilot.

Your requirements are tough, the careless  pilot scenario seems the easiest way to go :) There is, of course the possibility of the plane being fitted with ferry tanks, unless absolute realism is of importance.

@jpe61
I saw someone fly from Los Angeles to Hawaii on a Cessna 172, but the machine was totally overloaded and full of extra tanks. I have three more passengers to fly with the pilot.


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Minimum length required for the runway landing and takeoff of ATR 72-600?
What is the minimun length required for the runway landing and takeoff of ATR 72-600?
obs: altitude of 1000ft at 30 degrees

ATR42-600 and ATR72-600 require a different runway length

Take-off: 1315m; Landing: 915m
from: https://www.atr-aircraft.com/our-aircraft/atr-72-600/

That standard value is probably for sea level @ 15°C, not for 1000ft @ 30°C as the question asks.

You’re right. Think my answer is not correct. I guess there’s a table for numbers somewhere.


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what is the difference between RVR 200m and R 200m (LVO jeppesen landing minima)
sometimes JEPPESEN approach chart shows low visibility landing minima as RVR in meter or sometimes just R in meter. What is difference….

Please provide examples of what you're referring to; it's much easier to answer this sort of question, and the information is more meaningful to others, when we can all see these expressions in context. Thanks!

What is a "jeppesen"? Is it a place, "Jeppesen"? (Capitals matter!)

There are different cases.

The RVR is provided by a State authority, State is mentioned in the top left corner and R is used instead of RVR.

For minimums provided by EASA AOM
State is replaced by Std/State (Standard up to 2020), and RVR is used. But:

For separate minimums listings (like 10-9S pages)
RVR, CMV and VIS are abbreviated as “R”, “C” and
“V”.

Source

Since 2022, for EASA AOM:
RVR is also abbreviated by "R", both for landing and takeoff:

Straight-in landing minimums will be shown as RVR
with prefix “R”, as provided within the EASA tables.
A Visibility, prefixed "V", will only be charted if a
VIS value is published as State minimum. The prefix
“VIS” may still be used on older charts

Take -off minimums are shown as RVR “R”, as VIS
“V” or as RVR/VIS “R/V”. Values which could be
either RVR or VIS may be depicted without any prefix
on older charts.

A Visibility, prefixed "V", will only be charted for take-
off if a VIS value is published as State minimum.
For separate minimums listings (like 10-9S pages)
RVR, CMV and VIS are always abbreviated as “R”,
“C” and “V”.

Source.


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With flight dynamic equations, how do I return to positive alpha if alpha is always negative with negative w?
I'm coding a flight dynamics simulation, but my simulation runs into a problem with negative angles of attack. Once alpha is negative, it is impossible to get back to a state where the plane is generating positive vertical velocity.
At the start of each frame of the simulation we have some forward velocity ($u$) and vertical velocity ($w$), and at the end of each frame we calculate new values for $u$ and $w$. To keep things simple, assume the plane generates 0 lift at 0 alpha, we're always in level flight with no change to pitch or roll, and the pilot only makes changes to $u$ with the throttle.
If $u$ decreases enough that we generate less lift force than gravity, $w$ becomes negative and the plane begins to descend. The usual formula for angle of attack ($$) is a function of $u$ and $w$:
$ = arctan(w / u)$
This means a negative $w$ and positive $u$ will always give a negative $$.
I calculate lift with this equation (simplified. $q$ is dynamic pressure):
$L = 2π *  * q$
This means a negative $$ will always give negative lift.
And finally, the next frame's $w$ is just a function of lift and gravity (in level flight):
$w_{next} = w + L / mass - g$
So it is possible to have positive $$ and lift but end up with a negative $w$ due to gravity.
If a frame of my simulation ends with negative $w$, the simulation is put in a state where it is impossible to ever return to positive $$, because negative $w$ always gives negative $$. Negative $$ always gives negative $L$. And negative $L$ always gives negative $w_{next}$ Negative $w_{next}$ always gives negative $$ on the next frame.
Increasing $u$ eventually should return me to positive $L$ and $w$ as $q$ increases, but doesn't - it only generates more negative lift. What factors am I missing to be able to get back to positive $w$ with just an increase in $u$?

I think you simply have one wrong minus sign. Negative w means positive alpha.

Also you can only get a constant descent in this situation, you need to pitch up for level flight or climb. But that's a separate issue from correcting your equation.

My simulation accounts for a couple more factors like a separate zero-lift AoA so you can achieve level flight in this situation, but I simplified it a bit for the question

I think your problem is in part, that you assume that $w$ is positive if your aircraft is ascending. The opposite is the case. The aircraft coordinate system is defined as a right-handed system with the z-axis pointing downward.

This means a positive $w$ indicates that your aircraft is descending.
Therefore, if your plane keeps a constant velocity $u$, the body-fixed downward velocity $w$ will get positive because of the gravity pulling you down. This in turn results in a positive $\alpha$ providing lift. Eventually if you have implemented your simulation correctly, the $\alpha$-angle (and $w$) will oscillate for a bit at the beginning of your simulation, but after finding the steady-state, it will steadily descend with a positive angle-of-attack and a positive value of $w$.
P.S.: For the calculation of $\alpha$ it is very helpful to use the atan2() function, as it correctly discerns the various sign cases.

I don't think atan2 is warranted for alpha computations, any flight mechanics book I find has a simple atan and alpha is defined in the +- pi/2 range (consistent with atan)

@Federico atan2 is a broader implementation of atan, and has the same output in the +-pi/2 range as atan, therefore it is just as valid as the atan function. Also alpha is not only defined in this range, for example DIN9300 does not mention any constraints. However if you leave the realms of traditional fixed wing flight (VTOL, Helicopters, Multicopters etc.) with alpha values exceeding +-2pi, I promise you that the atan2() function will save your life safer :D.

I know very well what atan2 is. I still say that is too much for alpha, since any output outside of the +-pi/2 range given by atan2 are meaningless in my experience, even for rotorcraft (you don't have alpha ~= pi when flying backwards, it is still ~=0 for what concerns the aerodynamic equations)

@Federico no I dont think so. Image a profile in a VTOL application which gets blown at from the back. Or a Helicopter during a CAT-A start which entails flying backwards. In all of these applications you want to capture that the airflow is coming from the back. Dont you think?

yeah, ok, for craft with significantly asymmetric bodies it might make sense

This fixes a couple issues with my simulation, thanks. I knew that most models used -w for ascent and +w for descent, but I had flipped it since OpenGL is inverted where +y translation is up and -y translation is down (it also uses y-up instead of z-up). But I hadn't thought about the effect of that switch on AoA calculations. So now I only flip the sign when doing the final translation on the matrix, and it's improved a lot


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Why are Stratolaunch's engines so far forward?
Stratolaunch's engines hang from pylons, not under the wing, but entirely forward of the wing's leading edge.  (Even farther forward than, say, the 747's.)  What benefit does this have, to compensate for the heavier pylons?

Good question. I wonder if and how the center of gravity moves with the payload?

I bet the engine are placed as they are to achieve certain CoG. The CoG is the same regardless of payload. Otherwise this contraption would not fly either with, or without the spacethingy it carries.

Jpe61's guess would explain why they didn't just move the wing forward to where the engines are.  That would move the heavy spacethingy forward too.

Looks very similar to the B-52 engine placement to my naive eye! https://www.geaviation.com/military/engines/b-52

The forward location has two benefits:

Wing upwash is smaller, so the flow direction at the intake varies less with speed and altitude. This makes the intake more efficient and requires less robustness in its layout.
The forward location shifts the wing's center of gravity forward and gives it better flutter damping. How that works is explained here.

Wing torsion in combination with the two fuselages will produce a flutter mode where both fuselages pitch in opposite direction, torquing the center wing in the process. The slender wing in combination with the high pitch inertia of the fuselages will result in a very low flutter frequency. I suspect that damping this mode is of paramount importance for the Stratolaunch.
I also suspect that the high wing on a high fuselage allows more vertical distance between engine and wing, so the exhaust stream will pass below the lowered flaps. The high turbulence in the exhaust stream would put additional loads on the flaps; to avoid this, Boeing uses exhaust gates or a gap in its trailing edge flaps. Such a gap would not be needed with the lower engine position of the Stratolaunch. However, to place them lower does not require to shift them forward, too.

Neat, I'd never thought of the engine as a mass balance.  But these two benefits apply to many airplanes, while other airplanes rarely (ever?  B-58 Hustler?) mount engines so far forward.  What's special about Stratolaunch?

@CamilleGoudeseune I'm not familiar with the Stratolaunch, but I suspect the high aspect ratio gives it more torsional deformation / less torsional stiffness than what is usual on other airplanes. The slender mid-section in combination with the high inertia of the two fuselages will produce a very low antimetric torsion mode (where both fuselages pitch in opposite direction), and damping this would be helpful. I expanded the answer accordingly.

Yikes, one nose up and the other down, increasing in amplitude... "you will not go to space today."

Another reason is to reduce the shearing force on the engine pylon.
To understand this, consider that the engines are producing thrust, and the wings and fuselages are producing drag, so the engine pylon is pulled backwards at the wing mount-point and forwards at the engine mount point.
If the pylon were vertical then these two forces would be shearing forces against the mount points. By making the pylon close to horizontal in angle, the two forces are more closely opposed, hence the pylon undergoes mostly tension forces instead, which creates much less stress on the pylon and mount points.

@JamesMc: The forces don't change with the pylon angle, so the loads on the attachments points are unaffected. This concerns only the stresses inside the pylon structure. By using a properly designed truss, those will become tension and compression loads only.


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Why is the main landing gear of the Cessna Caravan placed so far forward?
Why is it that the rear landing gear of the Caravan is placed so far forward?
I understand that the center of gravity must lie somewhere in the middle between the nose gear and the  rear gear but wouldn't it be more convenient to have positioned it a little more towards the rear, giving it (to my opinion) more stability?
This would avoid the need of placing the support on the the tail of the aircraft when it is parked.

Image shows clearly the support.
I suppose that this would not compromise stability during flight.

possible duplicate of Why do planes always land on rear wheels instead of the nose wheels?

Not really. My question is about the spacing of the rear and nose landing gear on this specific aircraft.

Well you're asking about why the main gear is positioned where it is, and the answer is the center of gravity, which that question discusses.

I have no real idea apart from "that's what the designers decided", but one point is that a longer wheelbase would increase the turning radius on the ground. That would make ground operations a little more awkward in backcountry or 'bush' flying and Caravans are widely used in those environments.

@FabrizioMazzoni Give users some time to answer your question, before marking one as accepted. For example, both answers are missing that a backwards located landing gear would induce a forward moment, i.e., "smashing" your nose gear down uncontrollably when your main landing gear hits the ground.

Agree. Went to quickly for accepting

Wouldn't it be way cooler and faster a Grand Caravan fitted with the same landing gear system of a 210? Hopes and dreams I suppose.

You're right, placing the gear more aft would make ground handling much easier and not affect flight stability. But it would make taking off harder.
When the aircraft has accelerated to a speed just below lift-off speed, the wing's angle of attack is too low to create enough lift when that speed is reached, so the pilot will lift the nose while the main gear is still on the ground. This maneuver is called "rotation", and the speed at which the pilot pulls the stick or yoke back to lift up the nose is called rotation speed. By the time the aircraft has the right angle of attack, the lift-off speed is reached and the aircraft takes off.
If the gear is further aft, the downforce on the elevator needed to push the nose up needs to be bigger, and more drag is created during rotation. If the gear is still further aft, the maximum possible downforce at that speed would not be sufficient to get the nose up, and rotation must be delayed. The acceleration up to rotation speed is done with horizontal ground attitude because then the wing creates less lift and also less drag during the take-off run. Therefore, the pilot has to get the nose up to get the plane into the air, or he would need to accelerate to a much higher speed, at which the airplane flies even with its ground attitude. This all helps to minimize the airfield requirements.
To strike the right compromise between ground handling and elevator downforce during rotation, it is recommended to position the main gear along a line from the center of gravity down to the ground which is slanted 10°-15° back. The drawing below shows limits between 6° and 20°, but this is quite a wide range. Details depend on the desired angle of attack at liftoff and the maximum range of the center of gravity.

Side view of the recommended tricycle gear layout (Picture source).

The change of pitching moment on landing would also be hard to handle.

The rear landing gear in a tricycle style plane is placed close to the CG to minimize loads when landing and taking off. When you land an aircraft what you are really doing is transferring the load from the wings to the landing gear. If the CG is directly in line (longitudinally) with the gear, the gear will see the total mass of the aircraft. Any variation from this point will create a moment which will add to the load placed on the gear. Designers strive to minimize this.
On a similar note when a plane takes off it rotates about the axis of the wheels. Again the father off the CG is from the gear (longitudinally), the more force it may see on rotation (take off).
To answer your question directly, the load on an aircraft when its parked on the ground is not a use case designers care about. Its cheaper and (from an engineering stand point) simpler to stick a support under the rear of a parked plane than compromise the gear by moving its location. You should also keep in mind that looks are irrelevant, the engine of an aircraft is heavy and far forward thus it helps to balance out the somewhat empty tail section. For what its worth stability is mathematically calculable and quite quantifiable, it has little to do with looks.
If you want to know more about aircraft loading (and how its calculated) check out this document published by the FAA.